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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 14: Q. 41 (page 883)

In Problems 7鈥� 42, find each limit algebraically.
$\lim_{x 鈫� - 1} \frac{x^{3} + 2 x^{2} + x}{x^{4} + x^{3} + 2 x + 2} .$

Short Answer

Expert verified

The answer is 2.

Step by step solution

01

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

$f (x) = \frac{x^{3} + 2 x^{2} + x}{x^{4} + x^{3} + 2 x + 2}$

Put $x = - 1$ in the numerator we get,

$x^{3} + 2 x^{2} + x = - 1 + 2 - 1 = 0 .$

Put $x = - 1$ in the denominator we get,

$x^{4} + x^{3} + 2 x + 2 = 1 - 1 - 2 + 2 = 0 .$

Since both numerator and denominator gives 0 means they both have $x = - 1$ as their common root.

02

Step. 2 Factorizing

Numerator , $x^{3} + 2 x^{2} + x = x^{3} + x^{2} + x^{2} + x = x^{2} (x + 1) + x (x + 1) = (x^{2} + 1) (x + 1) .$

Denominator,role="math" localid="1647142064867" $x^{4} + x^{3} + 2 x + 2 = x^{3} (x + 1) + 2 (x + 1) = (x^{3} + 2) (x + 1) .$

So,

$\lim_{x 鈫� - 1} \frac{x^{3} + 2 x^{2} + x}{x^{4} + x^{3} + 2 x + 2} = \lim_{x 鈫� - 1} \frac{(x^{2} + 1) (x + 1)}{(x^{3} + 2) (x + 1)} = \lim_{x 鈫� - 1} \frac{x^{2} + 1}{x^{3} + 2} .$

Now we can put the limit directly.

03

Step. 3 Final calculation of the limit

$\lim_{x 鈫� - 1} \frac{x^{2} + 1}{x^{3} + 2} = \frac{1 + 1}{- 1 + 2} = \frac{2}{1} = 2 .$

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